Solve x^2-3x+11+x^2-4x+1=12x^2-2x+31

$$x ^ { 2 } - 3 x + 11 + x ^ { 2 } - 4 x + 1 = 12 x ^ { 2 } - 2 x + 31$$

Solve for x (complex solution)

$$x=-\frac{7\sqrt{15}i}{20}-\frac{1}{4}\approx -0.25-1.355544171i$$

$$x=\frac{7\sqrt{15}i}{20}-\frac{1}{4}\approx -0.25+1.355544171i$$

Steps Using the Quadratic Formula

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}-3x+11-4x+1=12x^{2}-2x+31$$

Combine -3x and -4x to get -7x.

$$2x^{2}-7x+11+1=12x^{2}-2x+31$$

Add 11 and 1 to get 12.

$$2x^{2}-7x+12=12x^{2}-2x+31$$

Subtract 12x^{2} from both sides.

$$2x^{2}-7x+12-12x^{2}=-2x+31$$

Combine 2x^{2} and -12x^{2} to get -10x^{2}.

$$-10x^{2}-7x+12=-2x+31$$

Add 2x to both sides.

$$-10x^{2}-7x+12+2x=31$$

Combine -7x and 2x to get -5x.

$$-10x^{2}-5x+12=31$$

Subtract 31 from both sides.

$$-10x^{2}-5x+12-31=0$$

Subtract 31 from 12 to get -19.

$$-10x^{2}-5x-19=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -10 for a, -5 for b, and -19 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-10\right)\left(-19\right)}}{2\left(-10\right)}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-10\right)\left(-19\right)}}{2\left(-10\right)}$$

Multiply -4 times -10.

$$x=\frac{-\left(-5\right)±\sqrt{25+40\left(-19\right)}}{2\left(-10\right)}$$

Multiply 40 times -19.

$$x=\frac{-\left(-5\right)±\sqrt{25-760}}{2\left(-10\right)}$$

Add 25 to -760.

$$x=\frac{-\left(-5\right)±\sqrt{-735}}{2\left(-10\right)}$$

Take the square root of -735.

$$x=\frac{-\left(-5\right)±7\sqrt{15}i}{2\left(-10\right)}$$

The opposite of -5 is 5.

$$x=\frac{5±7\sqrt{15}i}{2\left(-10\right)}$$

Multiply 2 times -10.

$$x=\frac{5±7\sqrt{15}i}{-20}$$

Now solve the equation x=\frac{5±7\sqrt{15}i}{-20} when ± is plus. Add 5 to 7i\sqrt{15}.

$$x=\frac{5+7\sqrt{15}i}{-20}$$

Divide 5+7i\sqrt{15} by -20.

$$x=-\frac{7\sqrt{15}i}{20}-\frac{1}{4}$$

Now solve the equation x=\frac{5±7\sqrt{15}i}{-20} when ± is minus. Subtract 7i\sqrt{15} from 5.

$$x=\frac{-7\sqrt{15}i+5}{-20}$$

Divide 5-7i\sqrt{15} by -20.

$$x=\frac{7\sqrt{15}i}{20}-\frac{1}{4}$$

The equation is now solved.

$$x=-\frac{7\sqrt{15}i}{20}-\frac{1}{4} x=\frac{7\sqrt{15}i}{20}-\frac{1}{4}$$

Steps for Completing the Square

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}-3x+11-4x+1=12x^{2}-2x+31$$

Combine -3x and -4x to get -7x.

$$2x^{2}-7x+11+1=12x^{2}-2x+31$$

Add 11 and 1 to get 12.

$$2x^{2}-7x+12=12x^{2}-2x+31$$

Subtract 12x^{2} from both sides.

$$2x^{2}-7x+12-12x^{2}=-2x+31$$

Combine 2x^{2} and -12x^{2} to get -10x^{2}.

$$-10x^{2}-7x+12=-2x+31$$

Add 2x to both sides.

$$-10x^{2}-7x+12+2x=31$$

Combine -7x and 2x to get -5x.

$$-10x^{2}-5x+12=31$$

Subtract 12 from both sides.

$$-10x^{2}-5x=31-12$$

Subtract 12 from 31 to get 19.

$$-10x^{2}-5x=19$$

Divide both sides by -10.

$$\frac{-10x^{2}-5x}{-10}=\frac{19}{-10}$$

Dividing by -10 undoes the multiplication by -10.

$$x^{2}+\left(-\frac{5}{-10}\right)x=\frac{19}{-10}$$

Reduce the fraction \frac{-5}{-10} to lowest terms by extracting and canceling out 5.

$$x^{2}+\frac{1}{2}x=\frac{19}{-10}$$

Divide 19 by -10.

$$x^{2}+\frac{1}{2}x=-\frac{19}{10}$$

Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{2}x+\left(\frac{1}{4}\right)^{2}=-\frac{19}{10}+\left(\frac{1}{4}\right)^{2}$$

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=-\frac{19}{10}+\frac{1}{16}$$

Add -\frac{19}{10} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=-\frac{147}{80}$$

Factor x^{2}+\frac{1}{2}x+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{4}\right)^{2}=-\frac{147}{80}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{-\frac{147}{80}}$$

Simplify.

$$x+\frac{1}{4}=\frac{7\sqrt{15}i}{20} x+\frac{1}{4}=-\frac{7\sqrt{15}i}{20}$$

Subtract \frac{1}{4} from both sides of the equation.

$$x=\frac{7\sqrt{15}i}{20}-\frac{1}{4} x=-\frac{7\sqrt{15}i}{20}-\frac{1}{4}$$